projectile motion [Physics]

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From: Sam on 14 Apr 2010 14:56

Hi all,

I've been experimenting with an online projectile-motion application.

Link to the app is here

http://mathdl.maa.org/mathDL/3/?pa=content&sa=viewDocument&nodeId=1027&pf=1

If you open the simulation and click on the "copyright" symbol you
will see how they have solved the system of equations.

What mystifies me is how they eliminated acceleration!

Their solution for initial velocity v is

v = 4 * (x-x0)
---------------------------------------------------
cos(w) * sqrt ((x-x0)tan(w)+(y0-y))

That is, when the cannon is aimed at angle w, it needs velocity v to
hit the target.

The question is, how did they arrive at this solution. Most of the
literature gives us equations for horizontal range, vertical height
reached at time t etc but every solution includes the acceleration due
to gravity.

If someone could shed some light on the algebra/physics that would be
great.

thanks,
Sam

From: Androcles on 14 Apr 2010 16:03

"Sam" <sam.n.seaborn(a)gmail.com> wrote in message
news:dfe72000-44fa-480c-9ff6-1426e3061435(a)w42g2000yqm.googlegroups.com...
>
>
> Hi all,
>
> I've been experimenting with an online projectile-motion application.
>
> Link to the app is here
>
> http://mathdl.maa.org/mathDL/3/?pa=content&sa=viewDocument&nodeId=1027&pf=1
>
> If you open the simulation and click on the "copyright" symbol you
> will see how they have solved the system of equations.
>
> What mystifies me is how they eliminated acceleration!
>
> Their solution for initial velocity v is
>
> v = 4 * (x-x0)
> ---------------------------------------------------
> cos(w) * sqrt ((x-x0)tan(w)+(y0-y))
>
> That is, when the cannon is aimed at angle w, it needs velocity v to
> hit the target.
>
> The question is, how did they arrive at this solution.

By breaking the problem down into smaller chunks.

> Most of the
> literature gives us equations for horizontal range, vertical height
> reached at time t etc but every solution includes the acceleration due
> to gravity.
>
> If someone could shed some light on the algebra/physics that would be
> great.
>

First thing you should do is work out what initial velocity is needed to
just reach the target vertically only, and take acceleration into account.
It just so happens (think about it) that this initial velocity is the same
as the terminal velocity would be if the projectile was dropped from
the target down to the gun.
If I drop a projectile from a height of h feet and it falls at g fps/s,
it will take t seconds to reach the ground and be travelling with
velocity v.

You must work out what v and t are.

Now if I shoot it upwards with velocity v it will reach height h in
the same time t (and then fall back, but we don't care about that).

Next include the horizontal motion (no acceleration here) so that it
takes the same time t to reach the target horizontally as it does
vertically.
Now you have two velocities, one vertical and one horizontal, and
you must calculate the angle needed.

From: OG on 14 Apr 2010 16:59

From: NoEinstein on 15 Apr 2010 12:53

On Apr 14, 2:56 pm, Sam <sam.n.seab...(a)gmail.com> wrote:
>
Dear Sam: Some of the first modern computers were designed,
specifically, to do the math on the ballistics of projectiles.
Amazingly, they got the results correct, while having the formula for
the acceleration due to velocity WRONG. The correct way to write
such is: g = 32.174 feet/per second EACH second (NOT per second^2!).
The velocity increases LINERALY, not parabolically! That humongous
error by Galileo and later by Newton is the likely reason that both
Coriolis and Einstein supposed, wrongly, that the energy or kinetic
energy of accelerating objects increases exponentially (sic). If any
of those men had simply realized, as I have, that velocity changes
LINERALY, then the KE (or E) could only be increasing linearly, too.
To do otherwise would be to violate the Law of the Conservation of
Energy.

The DISTANCE of fall increases by the square of the time of fall.
Thats because falling objects always have a COASTING carry-over
velocity from the previous second(s). So, most of the distance of
fall is due to that coasting carry-over, NOT due to an increasing
velocity input. A 500 pound projectile will have a 500 pound downward
force acting on it from the time such leaves the gun barrel until it
hits the target or the ground. Knowing that, plus the muzzle
velocity and the angle of fire, will allow calculating the
ballistics. None of those equations you cite are necessary if one
knows the (correct) simple physics that is in play. NoEinstein
>
> Hi all,
>
> I've been experimenting with an online projectile-motion application.
>
> Link to the app is here
>
> http://mathdl.maa.org/mathDL/3/?pa=content&sa=viewDocument&nodeId=102...
>
> If you open the simulation and click on the "copyright" symbol you
> will see how they have solved the system of equations.
>
> What mystifies me is how they eliminated acceleration!
>
> Their solution for initial velocity v is
>
> v = 4 * (x-x0)
> ---------------------------------------------------
> cos(w) * sqrt ((x-x0)tan(w)+(y0-y))
>
> That is, when the cannon is aimed at angle w, it needs velocity v to
> hit the target.
>
> The question is, how did they arrive at this solution. Most of the
> literature gives us equations for horizontal range, vertical height
> reached at time t etc but every solution includes the acceleration due
> to gravity.
>
> If someone could shed some light on the algebra/physics that would be
> great.
>
> thanks,
> Sam

From: Puppet_Sock on 16 Apr 2010 13:36

On Apr 14, 2:56 pm, Sam <sam.n.seab...(a)gmail.com> wrote:
[snip]
> What mystifies me is how they eliminated acceleration!
>
> Their solution for initial velocity v is
>
> v = 4 * (x-x0)
> ---------------------------------------------------
> cos(w) * sqrt ((x-x0)tan(w)+(y0-y))

The 4 should be sqrt(g). Probably copied their hand done
notes to the computer, and their hand notes were even worse
than my own. And nobody wants that.
Socks

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